8.4 Rotational Work and Kinetic Energy

Day 17 – June 17 – WBL 8.4-8.5

In the last 3 lectures, we have seen that there are numerous analogues between translational and rotational parameters. We continue along this line of reasoning with the study of rotational work. If a force acts in a tangential direction, moving an object along an arc length s, then the work done is

(as usual, must be expressed in radians). If this work is done over a time t, then the rotational power during this time is

In class, we will take a close look at what actually constitutes a “tangentially-directed force”.

8.4 Rotational Work and Kinetic Energy

PC141 Intersession 2013 Slide 1

Day 17 – June 17 – WBL 8.4-8.5

We know that any work done on a system will change its kinetic energy. Using the rotational form of Newton’s 2nd law (), we have

Assuming that torque (and therefore ) is constant, the rotational kinematic equations from chapter 7 tell us that , from which we find

But the work-KE theorem says that . This allows us to write an expression for rotational kinetic energy



Day 17 – June 17 – WBL 8.4-8.5

All that remains to be discussed in this section is the kinetic energy associated with an object that is rolling. Recall from the last lecture that rolling is a combination of translation (of the CM) and rotation (about the CM). The total kinetic energy involved in rolling is simply the sum of the translational and rotational kinetic energies; for an object of mass M,

Note that this equation is valid for any object that rolls without slipping (sphere, cylinder, etc.); they would simply have different . Often, you will encounter rolling problems in which the radius of the object isn’t provided. Don’t worry! Remember, is proportional to R2, but . Therefore, the radius cancels out of the K equation entirely.



Day 17 – June 17 – WBL 8.4-8.5

Problem #1: Bowling Ball


WBL LP 8.15A bowling ball rolls without slipping on a flat surface. The ball has…

A …rotational kinetic energy

B …translational kinetic energy

C …both translational and rotational kinetic energy

D …neither translational nor rotational kinetic energy

Day 17 – June 17 – WBL 8.4-8.5

Problem #2: Tipped Pencil


A pencil 18 cm long stands vertically on its point end on a horizontal table. If it falls over without slipping, with what tangential speed does the eraser end strike the table?

Solution: In class

WBL EX 8.55

Day 17 – June 17 – WBL 8.4-8.5

Problem #3: Loop-the-Loop Revisited


A steel ball rolls without slipping down an incline into a loop-the-loop of radius R. On the Day 14 lecture, we considered a sliding mass in the same situation, and calculated that it required a minimum speed of at the top of the loop in order to stay on the track. We then calculated (using energy principles) that the sliding mass

WBL EX 8.63

had to be released from a height h = 2.5R in order to achieve this minimum speed.

For the rolling ball, the required minimum speed is the same, but the required height of release is not. Calculate the new h.

Solution: In class

Day 17 – June 17 – WBL 8.4-8.5

In chapter 6, we introduced linear momentum, . A related quantity is angular momentum, . For a single particle, the magnitude of angular momentum is the product of the moment arm and the linear momentum:

It has SI units of kg·m2/s. For a rotating object, is simply equal to r, as the linear momentum of any particle in the object is tangential. Since each particle in the object has the same ω, we can write the total angular momentum of the object as

If we wish to take the vector nature of L into consideration, then

8.5 Angular Momentum


Day 17 – June 17 – WBL 8.4-8.5

Again referring to chapter 6, we know that a net force will alter a system’s total linear momentum: (this was an alternate form of Newton’s 2nd law). Likewise, by applying the rotational form of Newton’s 2nd law, we find that (considering only the magnitude),

In other words, the net torque on a system is equal to the time rate of change of the system’s angular momentum.



Day 17 – June 17 – WBL 8.4-8.5

Conservation of Angular MomentumIf the net torque on a system is zero, then its angular momentum must be constant. In that case, we can write

In some systems, a reconfiguring of the mass distribution (which changes the moment of inertia) will result in a change in the angular velocity. This effect can be demonstrated as shown in the figure below. A man spins slowly



in a chair while holding weights in his outstretched arms. When he brings the weights in toward his body (lowering the moment of inertia), his angular velocity increases.

Day 17 – June 17 – WBL 8.4-8.5

Figure skaters also exploit the conservation of angular momentum. They start to spin with their arms (and one leg) stretched outward. Bringing these limbs together along the rotation axis leads to a substantial increase in the rotational



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speed (up to 7 revolutions per second).

Day 17 – June 17 – WBL 8.4-8.5

The orbit of planets about the sun (or the moon about the Earth, etc.) provides a good example of a system where angular momentum about the central body is conserved. We know that L is conserved in this case because we can easily show that the net torque on the orbiting body is zero. This is because the gravitational force on the orbiting body is directed toward the central body – thus, and are anti-parallel, and their cross-product () is zero.Conservation of angular momentum in orbital mechanics can be used to prove Kepler’s 2nd law (the law of areas), although we won’t attempt to show that here.



Day 17 – June 17 – WBL 8.4-8.5

As a final example of angular momentum, consider the helicopter shown below. Before it leaves the ground, there is no angular momentum (nothing is moving at all). If there was only one rotor, conservation of L dictates that as it rotated, the body of the helicopter would have to rotate in the opposite direction (the torque required to turn the rotor is internal; there are no external torques). By using a second rotor with an oppositely-directed rotation, the angular momentum can be kept at zero without requiring the body to rotate.


Slide 12PC141 Intersession 2013

Day 17 – June 17 – WBL 8.4-8.5



The table shown here illustrates the many similarities between the translational and rotational dynamical equations.

Day 17 – June 17 – WBL 8.4-8.5

Problem #4: Angular Momentum Increase


WBL LP 8.19 (corrected)The angular momentum may be increased by…

A …decreasing the moment of inertia

B …decreasing the angular velocity

C …increasing the product of the angular speed and moment of inertia

D None of these

Day 17 – June 17 – WBL 8.4-8.5

Problem #5: Rotating Disk


A 10-kg rotating disk of radius 0.25 m has an angular momentum of 0.45 kg·m2/s. What is the angular speed of the disk?

Solution: In class

WBL EX 8.65

Day 17 – June 17 – WBL 8.4-8.5

Problem #6: Figure Skater


An ice skater spinning with outstretched arms has an angular speed of 4.0 rad/s. She tucks in her arms, decreasing her moment of inertia by 7.5%.

a) What is the resulting angular speed?b) By what factor does the skater’s kinetic energy change?c) Where does the extra kinetic energy come from?

Solution: In class

WBL EX 8.71

Day 17 – June 17 – WBL 8.4-8.5

Problem #7: Angular Momentum of an Airplane


A 1200 kg airplane is flying in a straight line at 80 m/s, 1.3 km above the ground. What is the magnitude of its angular momentum with respect to a point on the ground directly under the path of the plane?

Note: The purpose of this problem is to illustrate that the presence of a non-zero angular momentum does NOT imply that an object is rotating or moving in a curved path.

Solution: In class

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8.4 Rotational Work and Kinetic Energy